Entropy and the Second and Third Laws

Entropy (S) is a measure of the dispersal of energy and the number of microscopic arrangements available to a system; the second law states that the total entropy of the universe increases in any spontaneous process, and the third law states that the entropy of a perfect crystal is zero at 0 K. In brief: melting 1 mol of ice at 273.15 K increases its entropy by ΔS = ΔH/T = 6010 / 273.15 = 22.0 J/(mol·K).

Entropy explains why heat flows from hot to cold, why gases expand into a vacuum and why some endothermic processes still occur spontaneously. It is the quantity that, together with enthalpy, determines the direction of chemical change. This article explains entropy in terms of probability and heat, lists standard entropy values, shows how to predict and calculate entropy changes, and states the second and third laws.

What is entropy?

Entropy has two complementary definitions:

• Statistical: S = k_B ln W, where W is the number of microstates (ways of arranging particles and energy) that correspond to the observed state, and k_B = 1.381 × 10⁻²³ J/K. The more ways there are, the higher the entropy. This equation is carved on Boltzmann’s tombstone in Vienna.
• Thermodynamic: for a reversible process at constant temperature, ΔS = q_rev / T. Adding the same heat to a cold system increases its entropy more than adding it to a hot system.

Entropy is often described as “disorder”, but it is more accurate to talk about the dispersal of energy and the number of accessible states. Its unit is J/K, and for molar quantities J/(mol·K). Entropy is a state function.

What do the second and third laws say?

Second law: for a spontaneous process, ΔS_universe = ΔS_system + ΔS_surroundings > 0. For a reversible process the total is zero, and it can never be negative. It is the reason why heat cannot flow spontaneously from a cold body to a hot one and why no engine can convert heat completely into work.

Third law: the entropy of a perfect crystalline substance is zero at absolute zero. This gives a reference point, so unlike enthalpy, absolute entropies can be tabulated. The entropy of a substance at any temperature is found by adding up q/T for heating from 0 K, including the contributions of phase changes.

How can you predict the sign of ΔS?

• Phase changes: S(gas) ≫ S(liquid) > S(solid) for the same substance.
• Gas moles in reactions: if the number of moles of gas increases, ΔS is positive; if it decreases, ΔS is negative.
• Temperature: entropy increases when a substance is heated.
• Volume: a gas that expands has a higher entropy.
• Dissolving: the entropy generally increases when a solid dissolves, and decreases when a gas dissolves.
• Molecular complexity: larger molecules with more atoms have a higher entropy, because they have more ways to vibrate and rotate.

Standard molar entropies

The table shows the general pattern: solids have the lowest entropy, followed by liquids, with gases the highest. Hard, rigid solids such as diamond have particularly low values. Within gases, larger molecules have a higher entropy.

How do you calculate entropy changes?

Phase changes

At the melting point or boiling point, the phase change is reversible, and ΔS = ΔH / T.

• Melting of ice: ΔS_fus = 6010 J/mol / 273.15 K = +22.0 J/(mol·K).
• Vaporization of water: ΔS_vap = 40,700 J/mol / 373.15 K = +109 J/(mol·K).

The much larger increase for vaporization reflects the huge increase in the available volume and the number of microstates. For many nonpolar liquids, ΔS_vap is about 88 J/(mol·K) (Trouton’s rule), while hydrogen-bonded liquids such as water have higher values.

Chemical reactions

ΔS°rxn = Σ n S°(products) − Σ m S°(reactants)

For N₂(g) + 3H₂(g) → 2NH₃(g), with S°(NH₃) = 192.5 J/(mol·K): ΔS° = 2(192.5) − [191.6 + 3(130.7)] = 385.0 − 583.7 = −198.7 J/K. The sign is negative, as predicted, because 4 mol of gas are converted to 2 mol.

Isothermal expansion of a gas

For an ideal gas expanding from V₁ to V₂ at constant temperature, ΔS = nR ln(V₂/V₁). When 1.00 mol doubles its volume, ΔS = 8.314 × ln 2 = 5.76 J/K.

How do you find the entropy change of the universe?

The surroundings receive the heat that the system loses: ΔS_surr = −ΔH_sys / T (constant pressure). For the ammonia reaction above, ΔH = −92.2 kJ at 298 K.

1. ΔS_surr = +92,200 / 298 = +309 J/K.
2. ΔS_sys = −198.7 J/K.
3. ΔS_univ = 309 − 198.7 = +110 J/K.

The total is positive, so the reaction is spontaneous at 298 K, even though the entropy of the system itself decreases. The heat released increases the entropy of the surroundings by more than the system loses.

Who developed the concept of entropy?

Sadi Carnot analyzed the efficiency of heat engines in 1824. Rudolf Clausius introduced the quantity in 1865 and named it entropy, from the Greek for “transformation”, and he formulated the second law with the statement that the entropy of the universe tends toward a maximum. In 1877 Ludwig Boltzmann gave the statistical interpretation. The third law was formulated by Walther Nernst in 1906, and Max Planck later put it in the form that the entropy of a perfect crystal is zero at 0 K.

Which mistakes are common?

• Thinking that entropy of the system must always increase: only the total entropy of the universe must.
• Mixing up J and kJ: enthalpy is usually in kJ, but entropy in J/K.
• Assuming that elements have zero entropy: unlike enthalpy of formation, S° of elements is not zero.
• Equating entropy with messiness: think of the number of microstates.
• Forgetting the temperature in ΔH/T: it must be in kelvins.